Probability question

#1
I need a probability question answering as soon as possible and i though this might be the place to come.

There are 13 balls in a hat being randomly drawn out one at a time. There are 4 green balls, 2 red balls and 7 black balls.

A selection of 8 balls are randomly drawn. What is the probability that 3 of the green ball are drawn in the first 8 balls AND none of the red balls.

If you can answer this with some basic working I would REALLY appreciate it.
 

Dason

Ambassador to the humans
#2
Hi! :welcome: We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.
 
#3
I'm 36 and a director of my own business in the UK. I've taken a real world question and anonymised it so that people can't infer what i'm actually looking which is an issue of corruption in my business.

I'll give it a go though...

On drawing 1 ball the probability of it being a green ball is (4/13)*100. The probability of it NOT being a red ball is (11/13)*100.
On drawing a 2nd ball the probability is conditional on what happened in the 1st draw so i'm guessing it's some kind of series of multiplications of these probabilities.
 

Dason

Ambassador to the humans
#5
The answer isn't too hard to get but the little details you told us about your actual problem make me think that the probability question you provided won't actual get at the entire question you actually have. Of course I can't know that for sure without having more information about your *actual* question though. If you can't post it publicly maybe if you sent me a PM I could make it more anonymous for you if you want and help you along from there.
 

ted00

New Member
#6
The probability is 0.06526807, or about 6.5%

\(
Pr(\text{3 green, 0 red in 8 draws})
=Pr(\text{3 green, 0 red, 5 black})
=\frac{\binom{4}{3}\binom{2}{0}\binom{7}{5}}{\binom{13}{8}}
\)

Without having to derive the conditional probabilities we know this is the correct formula because the multivariate hypergeometric probability function is the formula to use when sampling without replacement from more than 2 groups. R code below.



Code:
# R code:
# multivariate hypergeometric for 
# 3 classes: green, red, black
# caps = population total
# small = sample
mh = function(G,R,B,g,r,b){
	choose(G,g)*choose(R,r)*choose(B,b)/choose(G+R+B,g+r+b)
}
mh(5,10,15,2,2,2) # wikipedia example -- 0.0795756
mh(4,2,7,3,0,5) # your example -- 0.06526807